Answer:
C. The 95% confidence interval is (0.2591, 0.3810). We are 95% confident that the true proportion of students attending the Spring Formal is between 25.91% and 38.10%.
Step-by-step explanation:
The interpretation of a confidence interval at a x% confidence level if that we are x% sure that the true proportion(mean) of the population is in this interval.
Confidence interval
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of ![1 - \frac{\alpha}{2}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B%5Calpha%7D%7B2%7D)
For this problem, we have that:
Seventy-two of the 225 students said they would attend the Spring Formal. So ![n = 225, \pi = \frac{72}{225} = 0.32](https://tex.z-dn.net/?f=n%20%3D%20225%2C%20%5Cpi%20%3D%20%5Cfrac%7B72%7D%7B225%7D%20%3D%200.32)
95% confidence interval
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 - 1.96\sqrt{\frac{0.32*0.68}{225}} = 0.259](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.32%20-%201.96%5Csqrt%7B%5Cfrac%7B0.32%2A0.68%7D%7B225%7D%7D%20%3D%200.259)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 + 1.96\sqrt{\frac{0.32*0.68}{225}} = 0.381](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.32%20%2B%201.96%5Csqrt%7B%5Cfrac%7B0.32%2A0.68%7D%7B225%7D%7D%20%3D%200.381)
The correct answer is:
C. The 95% confidence interval is (0.2591, 0.3810). We are 95% confident that the true proportion of students attending the Spring Formal is between 25.91% and 38.10%.